Sorry if this question is below the level of this site: I've read that the quotient of a Hausdorff topological group by a closed subgroup is again Hausdorff. I've thought about it but can't seem to figure out why. Is it obvious? A simple yes or no (with reference is possible) is all I need.
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2 Answers
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Edit: Below I expand my crude original answer "Yes" as requested by the community.
Yes. Let $G$ be the group and $H$ be the closed subgroup. The kernel of the quotient map $G \to G/H$ is equal to $\Delta^{-1}(H)$ where $\Delta : G \times G \to G$ is the continuous function $\Delta(x,y)= x- y$. Hence the kernel is closed. According to this $G/H$ is Hausdorff.
answered Dec 15, 2009 at 18:46
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$\begingroup$ Given that the result is false without group structure, and given that your link is to the wiki article rather than a particular subsection, -1. Will upvote if you expand on this $\endgroup$ Commented Dec 15, 2009 at 19:12
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$\begingroup$ I am not sure a question like this needs much more explanation. See the discussion at mathoverflow.net/questions/9014/field-structure-for-rn $\endgroup$ Commented Dec 15, 2009 at 19:21
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$\begingroup$ But then don't answer because this link is useless :) $\endgroup$– user717Commented Dec 15, 2009 at 19:32
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$\begingroup$ Yes, the last line certainly implies G/H Hausd iff H closed. However, why is the last line true. (My apologies is this glaringly obvious.) $\endgroup$ Commented Dec 15, 2009 at 19:46
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$\begingroup$ +1 (i.e. revert now we have expanded on this). As people have said, the key is that in a top group a weak separation axiom boosts up to an a priori stronger one. $\endgroup$ Commented Dec 15, 2009 at 20:32
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In fact, an even stronger statement holds: If $G$ is a topological group and $H$ is an (abstract) subgroup, then $G/H$ is Hausdorff if and only if $H$ is closed (cf Bourbaki, General Topology, III.2.5, prop 13). It's not hard to prove.
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$\begingroup$ Yes, before I posted the question I saw that H closed implies G/H T_0. What I couldn't see was that T_0 implies T_2 for a topological group. Is it obvious? $\endgroup$ Commented Dec 15, 2009 at 19:43